Problem: Suppose that f(x+3)=3x2+7x+4 and f(x)=ax2+bx+c. What is a+b+c?
Answer Choices:
A. −1
B. 0
C. 1
D. 2
E. 3
Solution:
Expanding, we have f(x+3)=a(x2+6x+9)+b(x+3)+c= ax2+(6a+b)x+(9a+3b+c). Equating coefficients implies that a=3,6⋅3+b=7, whence b=−11, and then 9⋅3+3⋅(−11)+c=4, and so c=10. Therefore a+b+c=3−11+10=2.
OR
Note that
f(x)=f((x−3)+3)=3(x−3)2+7(x−3)+4=3(x2−6x+9)+7x−21+4=3x2−11x+10
Therefore a=3,b=−11, and c=10, giving a+b+c=2.
OR
The sum a+b+c is f(1)=f(−2+3)=3(−2)2+7(−2)+4=2.
The problems on this page are the property of the MAA's American Mathematics Competitions